展开 (x-1)ln(x-1)

解答

展开

(x - 1) ln (x - 1)

x ln (x - 1) - ln (x - 1)

求解步骤

(x - 1) ln (x - 1)

= ln (x - 1) (x - 1)

使用分配律:

a (b - c) = ab - a c

a = ln (x - 1), b = x, c = 1

= ln (x - 1) x - ln (x - 1) \cdot 1

= x ln (x - 1) - 1 \cdot ln (x - 1)

乘以:

1 \cdot ln (x - 1) = ln (x - 1)

= x ln (x - 1) - ln (x - 1)

e x p an d \frac{d x}{( x + a ) ( x + b )}

e x p an d tan (x) (cot (x) - 3)

e x p an d \frac{( 4 ^{n} + 4 ^{n - 1} ) ^{2}}{4 ^{2 n - 2}}

s im pl i f y - 2 (1835 - 35)

e x p an d \frac{sin ^{2} ( α ) ( 1 + cos ( α ) )}{1 - cos ( α )}